Three-phase balanced wye-wye system problem

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Homework Statement

In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages and V_an = 120e^(i30°) V rms. The power absorbed by the load is 3435 W and the load impedance is 10 + j2 Ω. Find the two possible line impedance if the power generated by the source is 3774 W. Which line impedance is more likely to occur in an actual power transmission system?

Z_line = ?
Z_load = 10+j2 ohm
P_load = 3435 W
P_source = 3774 W
V_s = 120e^(i30°) V rms

Homework Equations

V_an = I_aA (10 + j2 + Z_line)

P_s = V_an * I_aA cos θ / sqrt(3)

The Attempt at a Solution

I can't seem to figure out a way to solve for any particular variable... I can find that the power of the line, P_line = 339 W, but that's about as far as I can get. The line impedance appears to be solved by:

Z_line = V_an - I_aA (Z_load) / I_aA
(derived from first relevant eq.)

But the current isn't given, and if I derive current I_aA from second relevant equation, then I don't know cos θ. The fact that there is a cos θ, it does make sense that there would be two valid line impedances.

Any thoughts?

 

[KataKoniK]

Hi there,

I can offer some guidance on how to approach this problem. First, let's start by writing out the equations we have and see if we can manipulate them to get the information we need.

From the first relevant equation, we have:

V_an = I_aA (10 + j2 + Z_line)

From the second relevant equation, we have:

P_s = V_an * I_aA cos θ / √3

Since we know the values for P_s, V_an, and Z_load, we can plug those in and solve for I_aA:

3774 = (120e^(i30°)) * I_aA * cos θ / √3

Solving for I_aA, we get:

I_aA = 3774 * √3 / (120e^(i30°) * cos θ)

Now, we can substitute this value for I_aA into our first equation and solve for Z_line:

Z_line = V_an - I_aA * (10 + j2)

Substituting in the value for I_aA, we get:

Z_line = V_an - (3774 * √3 / (120e^(i30°) * cos θ)) * (10 + j2)

We also know that the power absorbed by the load is 3435 W, so we can use this information to solve for cos θ:

3435 = (120e^(i30°)) * I_aA * cos θ

Solving for cos θ, we get:

cos θ = 3435 / (120e^(i30°) * I_aA)

Now we can substitute this value for cos θ into our equation for Z_line and solve for two possible values of Z_line:

Z_line = V_an - (3774 * √3 / (120e^(i30°) * (3435 / (120e^(i30°) * I_aA)))) * (10 + j2)

Simplifying, we get:

Z_line = V_an - (3774 * √3 / (3435)) * (10 + j2)

So, the two possible values for Z_line are:

Z_line1 = V_an - (3774 * √3 / (3435)) * (10 + j2)

Z_line2 = V_an + (3774 * √